Derivation of proper time in acceleration in SR

[starthaus]

I hope you understand that for example, \frac{\Delta x}{\Delta t} is not always the same as \frac{dx}{dt}

You can't be serious.

I did not do any derivations in post #38. I

You mean, you guessed

\Delta x=\frac{a t_i^2}{2}?

For good reason this is not part of the webpage you are citing ("The Relativistic Rocket"), it is total nonsense as I have showed . If you just guessed it without any derivation, try deriving it. This is all I was asking.

 

[yuiop]

I hope you understand that for example, \frac{\Delta x}{\Delta t} is not always the same as \frac{dx}{dt}

You can't be serious.

I am serious. Do you disagree?

 

[Austin0]

Problem statement:

Proper acceleration = 2c /s.

The x axes of S and S' are parallel to each other and the relative velocity of the frames and the acceleration and the velocity of the accelerated object is parallel to the x axes.

The velocity of accelerating object is v=0 at time t=0 and the object accelerates for 10 seconds as measured in frame S. We will consider two events, one at the start and at one at the end of the acceleration period.

Event 1:
(x1,t1) = (0,0)
(x1',t2') = (0,0)

Event 2:
(x2,t2) = (Δx,10)
(x2',t2') = (Δx', Δt')

\Delta x = (x_2-x_1) = x_2 \quad , \quad \Delta t = (t_2-t_1) = t_2
\Delta x' = (x_2'-x_1') = x_2' \quad , \quad \Delta t' = (t_2'-t_1') = t_2'

v is the final velocity of the accelerating object in frame S and is also the relative velocity of frame S' to frame S.

The instantaneous gamma factor at the terminal velocity in frame S

\gamma = \sqrt{1+(a't/c)^2} = \frac{1}{\sqrt{1-v^2/c^2}} = 20.025

Final coordinate velocity in S

v = \frac{a't}{\sqrt{1+(a't/c)^2}} = a't/\gamma = \frac{2*10}{20.025} = 0.99875c

Coordinate distance Δx traveled in frame S

\Delta x = (c^2/a)*(\sqrt{(1+(a \Delta t /c)^2)} -1) = (c^2/a)*(\gamma-1) = (1/2)*(20.025-1) = 9.5125

The origen of S' at (t=10) will be colocated with (t=10,x=9.9875) yes?
Considering that the accelerating system started out at v=0 while the S' system was moving at .99875c and the accl. Frame only attains that velocity at the end. how could the accl system end up having traveled almost an equal distance?
After 10 secs S' (.4994,0) will be at S (10,9.9875)
S(0.4994,0) will be located at S' (10,-9.9875) correct?
If this is correct it would seem to follow that event 2 must lie somewhere in the middle between these two events both spatially and temporally.

Coordinate distance Δx' in the CMIRF (S')

Fram the Lorentz transformation:

\Delta x' = \frac{\Delta x-v \Delta t}{\sqrt{1-v^2/c^2}} = \gamma(\Delta x-v \Delta t) = 20.025*(9.5125-0.99875*10) = -9.5125

You might find it surprising and unintuitive that the distance between the two events is the same in frame S and frame S', except for the sign. I know I did and maybe I made a mistake somewhere. .

A frame traveling 0.9521 relative to both S and S' would end up in the middle.
I.e. v= 0.9512 relative to S ,,,,and v= -0.9512 relative to S' but it could not have traveled 10 secs in either frame.

I may be totally wrong about all this but I think there may be a problem with using the CMIRF as I said last post.

Proper distance:

Distance is poorly defined in an accelerating reference frame. Proper distance in inertial RFs is normally the distance measured by a ruler between two simultaneous events. Since the two events are not simultaneous in any inertial reference frame in this example it is hard to define a proper distance even in the inertial reference frames. We can however invoke a notion of the invariant interval which can be thought of as the proper distance between events 1 and 2.

(c \Delta t)^2 - \Delta x^2 = (10)^2 - 9.5125^2 = 9.5125 .

Coordinate elapsed time Δt' in the CMIRF

Using the Lorentz transformation:

\Delta t&#039; = \frac{<br /> \Delta t-v \Delta x/c^2}{\sqrt{1-v^2/c^2}} = \gamma(\Delta t-v\Delta x/c^2) = 20.025*(10-0.99875*9.5125) = 10 s

Again this is a counter-intuitive result. .

Proper elapsed time:

As above ,,I think there may be a problem here

The proper elapsed time for the accelerating object is clearly defined because it measured by a single clock between the two events. It is derived like this. The total elapsed proper time is the integral of the instantaneous proper time at any instant which is a function of the instantaneous velocity u at any instant, so:

\frac{dt_0}{dt} = \frac{1}{\gamma^2} = \sqrt{1-u^2/c^2} = \frac{1}{\sqrt{1+(a_0\Delta t/c)^2}}

Integrating both sides with respect to t:

\Delta t_0 = \int \left( \frac{1}{\sqrt{1+(a_0t/c)^2}} \right) dt = (c/a_0)\, arsinh(a_0 \Delta t/c)

.

In this situation the elapsed time in S and S' do not really represent elapsed proper time but time as applied at different locations, so the clock desynchronization accounts for most of the difference between local clocks at event 2 and the accelerated systems elapsed proper time , yes?

 

[stevmg]

With all this discussion going on from my original thread I do NOT feel that I am such a novice after all!

stevmg

 

[yuiop]

Now it seems from playing around with my spreadsheet, that the following (without proof) is true:

\Delta x = \frac{1}{2} \alpha t_i^2

which has the same form as the Newtonian equation.

It obviously follows that the constant proper acceleration can be obtained from this simple equation:

\alpha = 2\frac{\Delta x}{t_i^2}

This cannot be since:

x=\frac{c^2}{a}(cosh(a \tau/c)-1)

dx=c d\tau sinh(a \tau/c)

You mean, you guessed

\Delta x=\frac{a t_i^2}{2}?

For good reason this is not part of the webpage you are citing ("The Relativistic Rocket") . If you just guessed it without any derivation, try deriving it. This is all I was asking.

When I first noticed the relationship, it "jumped out" at me when looking at the numbers on a spreadsheet and I did not not have a derivation, so yes in that sense I "guessed it". I now have the derivation as shown below:

Using notation defined thus:

Coordinate distance interval in S = \Delta x = d
Coordinate time interval in S = \Delta t = t
Proper acceleration = \alpha

.. the equation for coordinate distance from http://www.xs4all.nl/~johanw/PhysFAQ/Relativity/SR/rocket.html" is given as:

d = (c^2/\alpha)*(\sqrt{(1+(\alpha t /c)^2)} -1)

\rightarrow \alpha d = \sqrt{c^4+(\alpha t c)^2} - c^2

\rightarrow (\alpha d +c^2)^2 = c^4+(\alpha t c)^2

\rightarrow \alpha^2d^2 + c^4 + 2c^2\alpha d = c^4+\alpha^2 t^2 c^2

\rightarrow \alpha d^2 + 2c^2 d = \alpha t^2 c^2

\rightarrow 2c^2d =\alpha (t^2 c^2 - d^2)

\rightarrow \alpha = \frac{2d}{(t^2 - d^2/c^2)}


Since I defined t_i[/tex] as \sqrt{(t^2 - d^2/c^2) the following statents are true:<br /> <br /> \alpha = \frac{2d}{t_i^2}<br /> <br /> t_i = \sqrt{\frac{2d}{\alpha}}<br /> <br /> d = \frac{1}{2} \, \alpha t_i^2<br /> <br /> It is also an exact solution because I derived it directly from from an exact formula.<br /> <br /> Q.E.D.

 

[yuiop]

With all this discussion going on from my original thread I do NOT feel that I am such a novice after all!

stevmg

Well, acceleration in relativity is not a novice subject, but it is fairly straightforward.

 

[stevmg]

When I first noticed the relationship, it "jumped out" at me when looking at the numbers on a spreadsheet and I did not not have a derivation, so yes in that sense I "guessed it". I now have the derivation as shown below:

Using notation defined thus:

Coordinate distance interval in S = \Delta x = d
Coordinate time interval in S = \Delta t = t
Proper acceleration = \alpha

.. the equation for coordinate distance from http://www.xs4all.nl/~johanw/PhysFAQ/Relativity/SR/rocket.html" is given as:

d = (c^2/\alpha)*(\sqrt{(1+(\alpha t /c)^2)} -1)

\rightarrow \alpha d = \sqrt{c^4+(\alpha t c)^2} - c^2

\rightarrow (\alpha d +c^2)^2 = c^4+(\alpha t c)^2

\rightarrow \alpha^2d^2 + c^4 + 2c^2\alpha d = c^4+\alpha^2 t^2 c^2

\rightarrow \alpha d^2 + 2c^2 d = \alpha t^2 c^2

\rightarrow 2c^2d =\alpha (t^2 c^2 - d^2)

\rightarrow \alpha = \frac{2d}{(t^2 - d^2/c^2)}


Since I defined t_i[/tex] as \sqrt{(t^2 - d^2/c^2) the following statents are true:<br /> <br /> \alpha = \frac{2d}{t_i^2}<br /> <br /> t_i = \sqrt{\frac{2d}{\alpha}}<br /> <br /> d = \frac{1}{2} \, \alpha t_i^2<br /> <br /> It is also an exact solution because I derived it directly from from an exact formula.<br /> <br /> Q.E.D.

<br /> <br /> Beautiful!<br /> <br /> How did you get the &quot;quote within a quote&quot; as you did with this (your quoted post was quoted within starthaus&#039;s quote.)

 

[yuiop]

Beautiful!

How did you get the "quote within a quote" as you did with this (your quoted post was quoted within starthaus's quote.)

You can do it like this:

[quote="starthaus, post: 2873083"]

[quote="yuiop, post: 2873062"]
I hope you understand that for example,
  [tex]\frac{\Delta x}{\Delta t}[/tex] 
is not always the same as
 [tex]\frac{dx}{dt}[/tex] 
[/QUOTE]

You can't be serious.
[/QUOTE]

To get:

I hope you understand that for example, \frac{\Delta x}{\Delta t}
is not always the same as \frac{dx}{dt}

You can't be serious.

or:

[quote="yuiop, post: 2873062"]
I hope you understand that for example,
  [tex]\\frac{\Delta x}{\Delta t}[/tex] 
is not always the same as
 [tex]\frac{dx}{dt}[/tex] 

[quote="starthaus, post: 2873083"]
You can't be serious.
[/QUOTE]
[/QUOTE]

to get:

I hope you understand that for example, \frac{\Delta x}{\Delta t}
is not always the same as \frac{dx}{dt}

You can't be serious.

I wrapped it in PHP tags so that you can actually see the quote tags. Just use the multi quote feature and copy and paste to nest the quotes inside each other.

 

[stevmg]

You can do it like this:

To get:



or:

to get:





I wrapped it in PHP tags so that you can actually see the quote tags. Just use the multi quote feature and copy and paste to nest the quotes inside each other.

Thanks...

stevmg

 

[yuiop]

The origen of S' at (t=10) will be colocated with (t=10,x=9.9875) yes?

Yep, that makes sense if frame S' is traveling at 0.99875c and if the origins of S and S' were initially collocated.

Considering that the accelerating system started out at v=0 while the S' system was moving at .99875c and the accl. Frame only attains that velocity at the end. how could the accl system end up having traveled almost an equal distance?

The acceleration is extreme. In post #28 , I calculated that the accelerating object gets to nearly 0.8944c in the first second.

The accelerating object travels a distance of 9.5125 in 10 seconds so its average velocity in S is 0.95125c so it is not that much slower than the 0.99875c velocity of the CMIRF.

After 10 secs S' (.4994,0) will be at S (10,9.9875)
S(0.4994,0) will be located at S' (10,-9.9875) correct?
If this is correct it would seem to follow that event 2 must lie somewhere in the middle between these two events both spatially and temporally.

(t,x) = (10,9.5125) is event 2 in frame S and (t',x') = (10,-9.5125) is the same event 2 in frame S'. Where is this event 2 that lies somewhere between those events that you speak of?

You can't simply apply the gamma factor to the time in S to get the time in S' as 0.4994 because the initial vent and the final event are not at the same place in either reference frame. It is basically a simultaneity issue.

A frame traveling 0.9521 relative to both S and S' would end up in the middle.
I.e. v= 0.9512 relative to S ,,,,and v= -0.9512 relative to S' but it could not have traveled 10 secs in either frame.

The figure you have calculated for a reference frame that sees the origins of S and S' going away at equal speeds in opposite directions appears to be correct. This new frame would not appear to be in the middle in either frame S or S'. I am not sure why you think that is significant.

I may be totally wrong about all this but I think there may be a problem with using the CMIRF as I said last post.

All I can say is that using the CMIRF concept is a perfectly standard method in textbooks. Well at least I think it would be if I had any textbooks (it appears to be perfectly standard in serious online references anyway.)

In this situation the elapsed time in S and S' do not really represent elapsed proper time but time as applied at different locations, so the clock desynchronization accounts for most of the difference between local clocks at event 2 and the accelerated systems elapsed proper time , yes?

Correct again. That is why I have always referred to the elapsed times in S and S' as coordinate times rather than as proper times.

 

[starthaus]

When I first noticed the relationship, it "jumped out" at me when looking at the numbers on a spreadsheet and I did not not have a derivation, so yes in that sense I "guessed it".

Ok, I thought so.

\Delta x = d
\Delta t = t

Interesting choice of notation, so, for you:

\Delta x = x(t)-x(0)=x(t)

.. the equation for coordinate distance from http://www.xs4all.nl/~johanw/PhysFAQ/Relativity/SR/rocket.html" is given as:

d = (c^2/\alpha)*(\sqrt{(1+(\alpha t /c)^2)} -1)

\rightarrow \alpha d = \sqrt{c^4+(\alpha t c)^2} - c^2

\rightarrow (\alpha d +c^2)^2 = c^4+(\alpha t c)^2

\rightarrow \alpha^2d^2 + c^4 + 2c^2\alpha d = c^4+\alpha^2 t^2 c^2

\rightarrow \alpha d^2 + 2c^2 d = \alpha t^2 c^2

\rightarrow 2c^2d =\alpha (t^2 c^2 - d^2)

\rightarrow \alpha = \frac{2d}{(t^2 - d^2/c^2)}

So the above is nothing but

d = (c^2/\alpha)*(\sqrt{(1+(\alpha t /c)^2)} -1)

rearranged to express \alpha as a function of d with the notation t_i=\sqrt{(t^2 - d^2/c^2) . There is no physics in this, just an elementary algebraic manipulation.

 

[yuiop]

Coordinate distance interval in S = \Delta x = d
Coordinate time interval in S = \Delta t = t

Interesting choice of notation, so, for you:

\Delta x = x(t)-x(0)=x(t)

Yes it is a form of shorthand that is often used if the initial event is (0,0).
If event 1 is (0,0) and event 2 is (x,t) then \Delta x = x-0 = x and \Delta t = t-0 = t. Perfectly valid.

So the above is nothing but

d = (c^2/\alpha)*(\sqrt{(1+(\alpha t /c)^2)} -1)

rearranged to express \alpha as a function of d with the notation t_i=\sqrt{(t^2 - d^2/c^2) . There is no physics in this, just an elementary algebraic manipulation.

Yep, elementary and yet you could not see it and kept insisting it was wrong, until I showed you the proof. In the end all physics equations are an algebraic manipulation of something else.

The interesting physics aspect is that the proper time of an accelerating clock moving between two events is not the same as proper time of a clock with inertial motion between the same two events. Something you did not seem to realize, judging from your earlier postings. You seemed to think that the invariant proper time between two events, applied to accelerating clocks too.

 

[starthaus]

Yes it is a form of shorthand that is often used if the initial event is (0,0).
If event 1 is (0,0) and event 2 is (x,t) then \Delta x = x-0 = x and \Delta t = t-0 = t. Perfectly valid.

I understand now why you believe that dx and \Delta x are two different things.

Yep, elementary and yet you could not see it and kept insisting it was wrong, .

You are right, I couldn't see any physics , just numerology.

 

[Passionflower]

The interesting physics aspect is that the proper time of an accelerating clock moving between two events is not the same as proper time of a clock with inertial motion between the same two events.

And while we are at it, a traveler could have a special clock linked to an accelerometer that is calibrated at:

d \tau_{twin} = \frac{1}{2} \frac{\alpha}{c} \, \sinh \left(\frac{1}{2} \eta\right)

This clock would at each instant record the elapsed time of a hypothetical twin traveling between the same events but reaching him on a geodesic.

While another clock that shows the coordinate time has to be calibrated at:

d \tau_{coordinate} = \frac{c}{\alpha} \, \sinh \left(\eta)With Rapidity: \eta = \frac {\alpha d \tau}{c}

A more interesting question would be what would the above two formulas be for a positive or negative jerk (no slight intended). Anyone willing to take a stab at that one?

 

[Austin0]

The origen of S' at (t=10) will be colocated with (t=10,x=9.9875) yes?
Considering that the accelerating system started out at v=0 while the S' system was moving at .99875c and the accl. Frame only attains that velocity at the end. how could the accl system end up having traveled almost an equal distance?

Yep, that makes sense if frame S' is traveling at 0.99875c and if the origins of S and S' were initially collocated.
The acceleration is extreme. In post #28 , I calculated that the accelerating object gets to nearly 0.8944c in the first second.

The accelerating object travels a distance of 9.5125 in 10 seconds so its average velocity in S is 0.95125c so it is not that much slower than the 0.99875c velocity of the CMIRF.

The difference in total distance traveled is 0.4750 = [9.9875- 9.8512 ]
If the accl F reaches v=0.8944 in 1 sec
S' has traveld 0.99875 after 1 sec while acclF has only moved a percentage of ,8944 somewhat over 50% making a lead for S' that is already a large part of the final difference in dx',,, with a 0.10435 c velocity differential that will be reduced at a cubic falloff rate for acclF for 9 more secs.
It seems hard for the difference at the end to be so slight.

After 10 secs S' (.4994,0) will be at S (10,9.9875)
S(0.4994,0) will be located at S' (10,-9.9875) correct?
If this is correct it would seem to follow that event 2 must lie somewhere in the middle between these two events both spatially and temporally.

(t,x) = (10,9.9875) is event 2 in frame S and (t',x') = (10,-0.9875) is the same event 2 in frame S'. Where is this event 2 that lies somewhere between those events that you speak of?

The events I noted above are not two views of your event 2 at all.

They are event #3
Colocation S(10, 9.9875), S'(0.4994, 0)

And event #4
Colocation S'( 10, -9.9875) , S(0.4994, 0)

Unless I am mistaken your event #2 is:
The colocation of acclF at reaching v=0.99875 with S( 10, 9.512),S'( 10, -9.512)
Is this not correct?
DO you disagree that this event #2 lies between events 3 and 4 in both frames??

You can't simply apply the gamma factor to the time in S to get the time in S' as 0.4994 because the initial vent and the final event are not at the same place in either reference frame. It is basically a simultaneity issue.

Colocation S(10, 9.9875), S'(0.4994, 0),,,,OK you are telling me I am wrong here
,so please tell me what you think is the correct time for S' x'=0 cojacent with S t=10 ,x=9.9875?
ANd likewise for S t=0.4994 colocated with S' t'=10, x'= -9.9875.

A frame traveling 0.9521 relative to both S and S' would end up in the middle.
I.e. v= 0.9512 relative to S ,,,,and v= -0.9512 relative to S' but it could not have traveled 10 secs in either frame.

The figure you have calculated for a reference frame that sees the origins of S and S' going away at equal speeds in opposite directions appears to be correct. This new frame would not appear to be in the middle in either frame S or S'. I am not sure why you think that is significant.

You could say I found it an interesting coincidence that this v was exactly the average velocity you have calculated for asslF. And it may be significant.
You have the final distance from the origen and elapsed time the same in both frames for event 2.
Given that the origens clocks both read t,t'=0 and an understanding of simultaneity how do you think it is possible for clocks from both frames, colocated at a later point ,could agree on the proper time?
If your figures are correct and the end point of acceleration would be 9.512 in S because the acceleration was so rapid then this would seem to neccessarily imply an equally rapid deceleration relative to S'
This being the case how then could acclF end up traveling so far in S' i.e. -9.512 if the velocity differential dropped off so radically. From -0.99875 to -0.10435 in the first sec. Yes??
If on the other hand you assume that the deceleration in S' is the inverse of S
I.e. Starting out very slowly with a long term cubic increase in acceleration then that's fine but I think it would open a whole new can o' wormholes physicswise ,no?
Having frame agreement on profile between S and S' would seem problematic at best ,,for one , yeh??.
I suggest you may want to look at a drawing as far as colocating the event 2 spatial points.

Originally Posted by Austin0
it is based on acceleration relative to an abstract CMIRF and then this is transformed into rest frame coordinate acceleration at the end. This may of course be absolutely valid but it seems to me that in this circumstance the CIMRFs are somewhat of a bootstrap construct i.e. accelerating relative to one and then there is automatically another one there to accelerate from , with no direct connection to the observation from the reference frame , of either the acceleration of the actual system or the acceleration of the CMIRF.
The increased velocity is just assumed. The coordinate acceleration in the reference frame would actually have to be based on a series of short interval "instantaneous" velocity measurements , no?

I may be totally wrong about all this but I think there may be a problem with using the CMIRF as I said last post.

All I can say is that using the CMIRF concept is a perfectly standard method in textbooks. Well at least I think it would be if I had any textbooks (it appears to be perfectly standard in serious online references anyway.)

I don't doubt that and I am equally bereft of books, but what I am talking about is not a quantitative one , not about having infinitesimal measurements etc. Sporadic measurements would be fine. Its the fact that CMUFOs are not a matter of measurement in an inertial frame whatever. They are an ad hoc abstract creation without history or physics , simply a handy tool for the a priori assumtion of constant proper acceleration. It seems to me possible that they have the same problem to be found with accelerated lines of simultaneity which are simply another mainifestation of CMIRFs
I.e. they may have a questionable relation to the real world and its physics but that just MHO

SO at best they appear to be a superfluous addition ,easily eliminated simply by applying differential calculus directly to the accelerated frame or is there something I am missing?

In this situation the elapsed time in S and S' do not really represent elapsed proper time but time as applied at different locations, so the clock desynchronization accounts for most of the difference between local clocks at event 2 and the accelerated systems elapsed proper time , yes?

Correct again. That is why I have always referred to the elapsed times in S and S' as coordinate times rather than as proper times.

OK ,,,,,well then what is the completely inertial clock that shows greater elapsed proper time ?

 

[stevmg]

starthaus and yuiop:

There IS a difference between a "WAG" (wild a-- guess) and a "SWAG" (Scientific wild a-- guess.) This was an example of a SWAG.

SWAGs have more credibility than WAGs.

If we keep this straight, we're in business.

Doc
(stevmg)

 

[yuiop]

I understand now why you believe that dx and \Delta x are two different things.

Do you agree with the following statements:

When describing the motion of a particle with constant velocity dx = \Delta x .

When describing the motion of an accelerating particle dx \ne \Delta x .

I stated that dx and \Delta x are not necessarily the same thing, which is consistent with the above two statements.

You are right, I couldn't see any physics , just numerology.

This is just offensive You asked for a proof and I provided it for you and yet you still call it numerology.

When Kepler came up with his laws of orbital motion he analysed the data and discerned some patterns and came up with relationships that are now called his laws. Even though Kepler did not know the physics behind his laws (Newton did that), Kepler contribution is not thought of as insignificant and not many people would be so uncharitable as to describe Kepler as just a numerologist. I noticed a relationship and eventually provided a proof as well, so calling numerology is uncharitable. Analyzing data and discerning patterns is part of the scientific process.

In this thread I have provided a derivation for relativistic force and acceleration, a derivation for distance traveled by an accelerating object based on the relationship between force acceleration, momentum and energy and finally the equation for acceleration from the distance equation. I have also provided derivations for the terminal velocity achieved by an accelerating object and the proper time that elapses for the accelerating object. Saying that all you see is "numerology" (when you can not find any technical faults in my derivations) is just designed to be confrontational and against the rules of this forum. I suspect it is just sour grapes on your part, because you probably believed I would not be able to come up with a proof for \alpha = \frac{2d}{(t^2 - d^2/c^2)} <br />. (I now have a much simpler proof btw)

 

[stevmg]

"Doc" = stevmg = WAG
Starthaus and yiuop = SWAG
Definitive proof (now by yiuop & by starthaus) = true science

Guys, keep it cool! I'm the dummy, not you. If you want to criticize anyone for being unknowledgeable, MAKE IT ME. I can handle it. If that's as bad as I've been called in my life, trust me, I can handle it. When someone calls me an a--hole, my response is, "Is that the best you can do?" If it is worse, I take it as a compliment.

But, now, sports fans, to my new topic, "Derivation of Hyperbolic Equations from the Lorentz transformations." I bet you this gets interesting and heated, too.

Doc
stevmg

Go to this new thread:
https://www.physicsforums.com/showpost.php?p=2875235&postcount=1

 

[starthaus]

Do you agree with the following statements:

When describing the motion of a particle with constant velocity dx = \Delta x .

When describing the motion of an accelerating particle dx \ne \Delta x .

I stated that dx and \Delta x are not necessarily the same thing, which is consistent with the above two statements.

You are trying to make new inventions in differential calculus. In mainstream science, both dx and \Delta x mean x_i-x_{i-1}

 

[stevmg]

Go to this new thread:
https://www.physicsforums.com/showpost.php?p=2875235&postcount=1[/QUOTE]

Doc
stevmg

 

[yuiop]

The difference in total distance traveled is 0.4750 = [9.9875- 9.8512 ]

I think that should be 0.4750 = [9.9875- 9.5125]

If the accl F reaches v=0.8944 in 1 sec
S' has traveld 0.99875 after 1 sec while acclF has only moved a percentage of ,8944 somewhat over 50% making a lead for S' that is already a large part of the final difference in dx'

After 1 sec the accelerating object will have traveled 0.618 units so it has only fallen behind by a distance of 0.99875-0.618 = 0.3807 units. After that the speed differential is very small and getting smaller so it does not lose much more distance and only loses 0.475 units after 10 seconds.

,,, with a 0.10435 c velocity differential that will be reduced at a cubic falloff rate for acclF for 9 more secs.
It seems hard for the difference at the end to be so slight.

Seems OK to me.

The events I noted above are not two views of your event 2 at all.

They are event #3
Colocation S(10, 9.9875), S'(0.4994, 0)

And event #4
Colocation S'( 10, -9.9875) , S(0.4994, 0)

Unless I am mistaken your event #2 is:
The colocation of acclF at reaching v=0.99875 with S( 10, 9.512),S'( 10, -9.512)
Is this not correct?
DO you disagree that this event #2 lies between events 3 and 4 in both frames??

O.K. you threw me off balance by introducing two new events to an already complicated situation! Yes, your definition of my event #2 is correct for the location and time that the accelerating object reaches it terminal velocity after 10 seconds in S and I have corrected the original post to reflect that. Your Lorentz transformations of the other events are also correct, but I am not sure what you are trying to demonstrate here.

Your event #3 = S(10, 9.9875) = S'(0.4994, 0), is where the CMIRF origin is after 10 seconds in S.

Your event #4 = S'( 10, -9.9875) = S(0.4994, 0) is where the origin of S is after 10 seconds in S'.

I wouldn't say that event 2 is half way between events 3 and 4 in either S or S' spatially or temporally. 0.4994 is approximately 1/20 of the time measured in the other frame. Events 2 and 3 but not 4 are simultaneous in S and events 2 and 4 but not 3 are simultaneous in S'.

You could say I found it an interesting coincidence that this v was exactly the average velocity you have calculated for asslF. And it may be significant.

I agree.

You have the final distance from the origen and elapsed time the same in both frames for event 2.
Given that the origens clocks both read t,t'=0 and an understanding of simultaneity how do you think it is possible for clocks from both frames, colocated at a later point ,could agree on the proper time?

That is what comes out of the Lorentz transformations and I showed the calculations in post #20 of this thread. I also said it is surprising and unintuitive, but that is what comes out of the maths and it unique to constant acceleration.

If your figures are correct and the end point of acceleration would be 9.512 in S because the acceleration was so rapid then this would seem to neccessarily imply an equally rapid deceleration relative to S'
This being the case how then could acclF end up traveling so far in S' i.e. -9.512 if the velocity differential dropped off so radically. From -0.99875 to -0.10435 in the first sec. Yes??
If on the other hand you assume that the deceleration in S' is the inverse of S
I.e. Starting out very slowly with a long term cubic increase in acceleration then that's fine but I think it would open a whole new can o' wormholes physicswise ,no?

The second case is correct (the can of wormholes).

Having frame agreement on profile between S and S' would seem problematic at best ,,for one , yeh??.
I suggest you may want to look at a drawing as far as colocating the event 2 spatial points.

A drawing might help us both. Unfortunately, all the velocities are near 0.9c in the chosen example and it is hard to see anything interesting clearly in the diagram. If you are really interested I can run all the numbers for an acceleration of 0.1c per second which results in a final CMIRF velocity of about 0.7c and a average velocity of about 0.4c for the accelerating object after 10 seconds. The lines on the diagram would be more spread out then.

I don't doubt that and I am equally bereft of books, but what I am talking about is not a quantitative one , not about having infinitesimal measurements etc. Sporadic measurements would be fine. Its the fact that CMUFOs are not a matter of measurement in an inertial frame whatever. They are an ad hoc abstract creation without history or physics , simply a handy tool for the a priori assumtion of constant proper acceleration. It seems to me possible that they have the same problem to be found with accelerated lines of simultaneity which are simply another mainifestation of CMIRFs
I.e. they may have a questionable relation to the real world and its physics but that just MHO

SO at best they appear to be a superfluous addition ,easily eliminated simply by applying differential calculus directly to the accelerated frame or is there something I am missing?

O.K. I take it you don't like CMIRFs . They are handy (to me anyway), because when you carry out the integration of let's say the instantaneous gamma factor to obtain the total elapsed proper time, it the CMIRF concept that allows you to justify the assumption that there is no additional time dilation due to acceleration per se.

OK ,,,,,well then what is the completely inertial clock that shows greater elapsed proper time ?

It is a clock traveling at a constant velocity equal to the average velocity of the accelerating clock. If this inertial clock starts out at the same time as the accelerating clock from location d1=0 in S they both end up at location d2 = 9.5125 in S at the same time.

 

[Passionflower]

I think that should be 0.4750 = [9.9875- 9.5125]
It is a clock traveling at a constant velocity equal to the average velocity of the accelerating clock. If this inertial clock starts out at the same time as the accelerating clock from location d1=0 in S they both end up at location d2 = 9.5125 in S at the same time.

Indeed, a loop, a twin experiment showing differential aging.

 

[Austin0]

Unless I am mistaken your event #2 is:
The colocation of acclF at reaching v=0.99875 with S( 10, 9.512),S'( 10, -9.512)
Is this not correct?
DO you disagree that this event #2 lies between events 3 and 4 in both frames??

Yes, your definition of my event #2 is correct for the location and time that the accelerating object reaches it terminal velocity after 10 seconds in S
Your Lorentz transformations of the other events are also correct, but I am not sure what you are trying to demonstrate here.
Not trying to demonstrate simply trying provide a context and boundaries to see how these, as you said , surprising results work out.
Your event #3 = S(10, 9.9875) = S'(0.4994, 0), is where the CMIRF origin is after 10 seconds in S.

Your event #4 = S'( 10, -9.9875) = S(0.4994, 0) is where the origin of S is after 10 seconds in S'.

I wouldn't say that event 2 is half way between events 3 and 4 in either S or S' spatially or temporally. 0.4994 is approximately 1/20 of the time measured in the other frame. Events 2 and 3 but not 4 are simultaneous in S and events 2 and 4 but not 3 are simultaneous in S'.

You could say I found it an interesting coincidence that this v was exactly the average velocity you have calculated for asslF. And it may be significant.

I agree.

If the events are viewed from the frame M traveling 0.9512c wrt each, it turns out that event #2 does occur at the midpoint between S origen and S' origen where each are 9.512 away from that event In that frame they are obviously separated by 19.24 at that time , or so it appears , According to their owns metrics. In M that distance would be reduced by a gamma of 3.2407.
I am still looking at this picture to see if it makes things any less surprising

You have the final distance from the origen and elapsed time the same in both frames for event 2.
Given that the origens clocks both read t,t'=0 and an understanding of simultaneity how do you think it is possible for clocks from both frames, colocated at a later point ,could agree on the proper time?

That is what comes out of the Lorentz transformations and I showed the calculations in post #20 of this thread. I also said it is surprising and unintuitive, but that is what comes out of the maths and it unique to constant acceleration.

Last night I remembered and realized I was mistaken about simultaneity and the clocks agreeing.
IF two frames are synched at the coincidence of the origens then all other locations are out of synch between them with one exception. There is a singularity. A static point equidistant between the origens where the colocated clocks from both frames will continue to agree on proper time as they pass each other.
In this case, that point coincides with the origen of frame M so the passing clocks of S and S' will always have the same time reading there.
In this frame the accelFrame will first be decelerating and then accelerating but reaching +0.9512c at the origen of M at S t=10,S' t'=10

If on the other hand you assume that the deceleration in S' is the inverse of S
I.e. Starting out very slowly with a long term cubic increase in acceleration then that's fine but I think it would open a whole new can o' wormholes physicswise ,no?

The second case is correct (the can of wormholes).

Well this presents an interesting study. In one frame a rapid increase in velocity that falls off over increasing distance and time is measured in another frame as a slow decrease that augments over a decreasing distance and time ,over the same course of acceleration. Yes?
Given that both measuring frames are inertial and uniform doesn't it seem hard to picture how they could agree on the short interval changes of velocity ?
The reciprocal acceleration profiles this requires

A drawing might help us both. Unfortunately, all the velocities are near 0.9c in the chosen example and it is hard to see anything interesting clearly in the diagram. If you are really interested I can run all the numbers for an acceleration of 0.1c per second which results in a final CMIRF velocity of about 0.7c and a average velocity of about 0.4c for the accelerating object after 10 seconds. The lines on the diagram would be more spread out then.

What would be interesting would be the acceleration profiles for an acceleration to 0.7 in one frame and the reciprocal deceleration in the 0.7 frame.
To be comprehensible it would seem to need spot values for each frame , maybe every 0.05 c and the appropriate locations and times for both frames. I worked on something like that previously but was working pn a wrong profile. I figured a single gamma dropoff.

O.K. I take it you don't like CMIRFs . They are handy (to me anyway), because when you carry out the integration of let's say the instantaneous gamma factor to obtain the total elapsed proper time, it the CMIRF concept that allows you to justify the assumption that there is no additional time dilation due to acceleration per se.

Yep , see a handy concept for justification but no actual use
Isn't the clock hypotheses enough to justify your assumption.
If that assumption is wrong there won't be any CMIRF around to apologize. Besides I thought that question had been fairly well , empirically answered??
It's not really that I don't like the critters as I haven't actually seem enough of em around to have an opinion.

 

[Passionflower]

And while we are at it, a traveler could have a special clock linked to an accelerometer that is calibrated at:

d \tau_{twin} = \frac{1}{2} \frac{\alpha}{c} \, \sinh \left(\frac{1}{2} \eta\right)

This clock would at each instant record the elapsed time of a hypothetical twin traveling between the same events but reaching him on a geodesic.

While another clock that shows the coordinate time has to be calibrated at:

d \tau_{coordinate} = \frac{c}{\alpha} \, \sinh \left(\eta)With Rapidity: \eta = \frac {\alpha d \tau}{c}

A more interesting question would be what would the above two formulas be for a positive or negative jerk (no slight intended). Anyone willing to take a stab at that one?

Oops, I just realized I made a mistake entering the first formula, here is the complete posting but now hopefully correct.
Sorry for that.

With Rapidity: \eta = \frac {\alpha \tau}{c}

Hypothetical inertial twin:

d \tau_{twin} = 2\frac{c}{\alpha} \sinh \left(\frac{1}{2} d\eta\right)

Coordinate time:

d \tau_{coordinate} = \frac{c}{a} \sinh(d \eta) = d \tau_{twin} \cosh \left(\frac{1}{2} d\eta\right)

 

[Anamitra]

To yuiop:

Consider an object of mass m₀ which, subjected to a constant force, accelerates at a₀ initially. Initially, the velocity of this mass is zero but then picks up as this force is applied.

By the relativistic momentum equation,

a = dv/dt = a₀\sqrt{(1 - v^2/c^2)}

dv/\sqrt{(c^2 - v^2)} = a₀dt/c

sin^-1 (v/c) = a₀t/c

v/c = sin (a₀t/c)

This is a periodic function (up and down depending on t) which is impossible

{a_{0}} has been assumed to be a constant[wrt time] in the process of integration . But this is raising a contradiction in the relation

{\frac{dv}{dt}{=}{a_{0}}{\sqrt{(1-v^2/c^2)}}

If the left side is positive increasing the right hand side should also increase.But velocity should increase due to the acceleration and hence {\sqrt{(1-v^2/c^2)}}decreases. This disturbs the balance of the above equation.
Again if the left side is positive constant the right side decreases due to increase of v.The balance gets disturbed.

Should the left side be negative and increasing in magnitude there is a chance of compatibility since v would decrease on the right side[if {a_{0}} is a negative constant in this case].But if initial speed is zero then this won't work.Incidentally v=0 at t=0 has been assumed in the calculations.

 

[stevmg]

The standard interpretation follows correctly from the INVARIANCE of the Minkowski metric as follows:

(c d \tau)^2=(cdt)^2-dx^2

starthaus

The above is a hyperbolic relationship between (cdt)^2 \text {and } dx^2

This is not the same as the hyperbolic relationship between
(ct)^2 \text {and } x^2
which also does exist.
What is the relationship between the two equations? One is not the derivative of the other. I am trying to figure out what
(ct)^2 - x^2 = a^2
means. These are a family of hyperbolas with a parameter
a
but what does each curve represent? What does
a
represent?

 

[starthaus]

starthaus

The above is a hyperbolic relationship between (cdt)^2 \text {and } dx^2

This is not the same as the hyperbolic relationship between
(ct)^2 \text {and } x^2
which also does exist.
What is the relationship between the two equations? One is not the derivative of the other. I am trying to figure out what
(ct)^2 - x^2 = a^2
means. These are a family of hyperbolas with a parameter
a
but what does each curve represent? What does
a
represent?

You can find all the answers https://www.physicsforums.com/blog.php?b=1911 . a=\frac{c^2}{a_p} where a_p represents the proper acceleration.

 

[stevmg]

Referring to post 77, I have downloaded the second .pdf (Lorentz...& Force in SR) as I had already downloaded the first months ago before I could make any attempt at comprehending it. Thanks for the clarification about a.

Now this leads to a "correlative" question about the answer in which a is the proper acceleration in that equation (ct)^2 - x^2 = a^2. Your answer would imply that any FR with any v and a = 0 would would be on the hyperbola (ct)^2 - x^2 = 0 which maps out to be the two diagonal asymptotes x = \pm (ct) for the general equation of the hyperbola (ct)^2 - x^2 = a^2.

In other words, just working with the positives, for given v, then, if x = vt then a = 0.

If this last paragraph is true, I am going to have to "digest" that to understand what that means.

Also, with proper times themselves, in non-accelerating FRs proper time represents the time that two events are apart given no displacement of x over t (i.e, forcing v = 0) and using Lorentz to show that \tau = \Delta t which is obvious (used to oblivious to me.)

But given that non-trivial acceleration \Rightarrow displacement of x then I don't know what the hell proper time means in that case.

 

[starthaus]

Referring to post 77, I have downloaded the second .pdf (Lorentz...& Force in SR) as I had already downloaded the first months ago before I could make any attempt at comprehending it. Thanks for the clarification about a.

Now this leads to a "correlative" question about the answer in which a is the proper acceleration in that equation (ct)^2 - x^2 = a^2. Your answer would imply that any FR with any v and a = 0 would would be on the hyperbola (ct)^2 - x^2 = 0 which maps out to be the two diagonal asymptotes x = \pm (ct) for the general equation of the hyperbola (ct)^2 - x^2 = a^2.

In other words, just working with the positives, for given v, then, if x = vt then a = 0.

If this last paragraph is true, I am going to have to "digest" that to understand what that means.

Also, with proper times themselves, in non-accelerating FRs proper time represents the time that two events are apart given no displacement of x over t (i.e, forcing v = 0) and using Lorentz to show that \tau = \Delta t which is obvious (used to oblivious to me.)

But given that non-trivial acceleration \Rightarrow displacement of x then I don't know what the hell proper time means in that case.

Let's look together at the first file, https://www.physicsforums.com/blog.php?b=1911 . Eq (15) tells you that :

x=\frac{c^2}{a_p}cosh\frac{a_p \tau}{c}
t=\frac{c}{a_p}sinh\frac{a_p \tau}{c}

so -(ct)^2+x^2=(\frac{c^2}{a_p})^2

What does this tell you?

Further, differentiating eq (15):

dx=c*sinh\frac{a_p \tau}{c} *d \tau
dt=cosh \frac{a_p \tau}{c}* d \tau

meaning that :

(cdt)^2-dx^2=(c* d \tau)^2

What does this mean?

 

[stevmg]

Tried to answer but window locked up too many times. Will try again tomorrow, 9-22.

 

[stevmg]

starthaus -

I have broken your post up into smaller segments and there will be sequential posts by me. This will prevent my computer from locking up. I am assuming that S and S' are constructed such that x = x&#039; = 0 and t = t&#039; = 0 at the beginning.

x=\frac{c^2}{a_p}cosh\frac{a_p \tau}{c}
t=\frac{c}{a_p}sinh\frac{a_p \tau}{c}

so -(ct)^2+x^2=(\frac{c^2}{a_p})^2

What does this tell you?

These equations (15) according to your article are the equations of hyperbolic motion. This is proven by the last equation in your quote which are in the form of a hyperbola with variables x \text { and } ct.

 

[stevmg]

Moving right along,

Further, differentiating eq (15):

dx=c*sinh\frac{a_p \tau}{c} *d \tau
dt=cosh \frac{a_p \tau}{c}* d \tau

meaning that :

(cdt)^2-dx^2=(c* d \tau)^2

What does this mean?

This establishes a hyperbolic equation on the differentials
(cdt) text [ and ] dx [/iex]

The parameter here is (c* d \tau)^2

 

[starthaus]

starthaus -

I have broken your post up into smaller segments and there will be sequential posts by me. This will prevent my computer from locking up. I am assuming that S and S' are constructed such that x = x&#039; = 0 and t = t&#039; = 0 at the beginning.



These equations (15) according to your article are the equations of hyperbolic motion. This is proven by the last equation in your quote which are in the form of a hyperbola with variables x \text { and } ct.

Good. What happens when the proper acceleration a_p increases?

 

[starthaus]

Moving right along,
This establishes a hyperbolic equation on the differentials
(cdt) text [ and ] dx [/iex]

The parameter here is (c* d \tau)^2

No, this gives you the relationship between proper (\tau) and coordinate (t) time:

d \tau =\sqrt{1-(\frac{dx}{dt})^2}*dt

used in the twins paradox

 

[stevmg]

Good. What happens when the proper acceleration a_p increases?

as a_p increases, (c/a_p)^2 gets smaller (\Rightarrow 0.) There is no upper limit on a_p as there is with v\Rightarrow c provided the v achieved is &lt;c.

 

[stevmg]

No, this gives you the relationship between proper (\tau) and coordinate (t) time:

d \tau =\sqrt{1-(\frac{dx}{dt})^2}*dt

used in the twins paradox

Of course. What the hell am I thinking.

The equation (cdt)^2-dx^2=(c* d \tau)^2 is not a hyperbola as none of those variables cdt, dx \text { or } c*d \tau is a constant. Where is my damn brain?

Now, I have seen c \Delta t \text { and/or } \Delta x used in the past when your equation above suggests cdt \text { or } dx. When is that appropriate?

 

[stevmg]

Since a_p = c*a (Equation 12), then the "constant" side of the hyperbolic equation is (\frac{c^2}{a_p})^2 \text { or } (\frac{c}{a})^2.

As a \Rightarrow \infty that constant portion \Rightarrow 0

As a \Rightarrow 0 that constant portion \Rightarrow \infty

That makes no sense.

By the Minkowski equations, (ct)^2 - x^2 = (\text {some constant})^2 even at a constant velocity. I am missing something.

 

[starthaus]

Since a_p = c*a (Equation 12), then the "constant" side of the hyperbolic equation is (\frac{c^2}{a_p})^2 \text { or } (\frac{c}{a})^2.

As a \Rightarrow \infty that constant portion \Rightarrow 0

As a \Rightarrow 0 that constant portion \Rightarrow \infty

That makes no sense.

By the Minkowski equations, (ct)^2 - x^2 = (\text {some constant})^2 even at a constant velocity. I am missing something.

Think again, what happens for a_p-&gt;\infty?

 

[JesseM]

But given that non-trivial acceleration \Rightarrow displacement of x then I don't know what the hell proper time means in that case.

Physically the "proper time" along a given worldline just means the time elapsed on a clock that travels along that worldline. If you want to know the proper time between two events on the worldline of a clock moving at constant velocity along the x-axis of some frame, it's just \Delta \tau = \sqrt{\Delta t^2 - (1/c^2) \Delta x^2 }, where \Delta t is the difference in coordinate time between the two events and \Delta x is the difference in coordinate position. This can be rewritten as \Delta \tau = \Delta t \sqrt{1 - (1/c^2) (\Delta x / \Delta t )^2 } = \Delta t \sqrt{1 - v^2 /c^2 } which is just the standard time dilation formula. For a worldline where the velocity is varying in a continuous way, you can approximate its worldline by a "polygonal" worldline made up of a series of constant-velocity segments each lasting a time interval of \Delta t with instantaneous accelerations between them, then if the velocity during the first time interval v₁ (which could just be the average velocity during the same interval for the path with continuously changing velocity that you are trying to approximate), the second has velocity v₂, and the final segment has velocity v_N, then the total elapsed time would just be the sum of the elapsed time on each segment, or \Delta t \sqrt{1 - v_1^2 /c^2} + \Delta t \sqrt{1 - v_2^2 / c^2} + ... + \Delta t \sqrt{1 - v_N^2 /c^2}. And in the limit as as the time interval of each segment goes to zero (so the number of segments approaches infinity), this approximation should approach perfect agreement with the proper time on the original path with continuously-changing velocity. Since an integral is just the limiting case of a sum with smaller and smaller intervals (or an infinite series of intervals which each have an 'infinitesimal' time dt), that means the proper time along a worldline where the velocity as a function of time is given by v(t) can always be computed according to the integral \int_{t_0}^{t_1} \sqrt{1 - v(t)^2 /c^2} \, dt

 

[yuiop]

Of course. What the hell am I thinking.

The equation (cdt)^2-dx^2=(c* d \tau)^2 is not a hyperbola as none of those variables cdt, dx \text { or } c*d \tau is a constant. Where is my damn brain?

Now, I have seen c \Delta t \text { and/or } \Delta x used in the past when your equation above suggests cdt \text { or } dx. When is that appropriate?

You can make any of those variables into a constant. If you draw a line from the origin that intercepts the hyperbola \Delta t = \sqrt{(3^2 + \Delta x^2)} then for ANY velocity \Delta x/ \Delta t in the range (-c<v<c) the elapsed proper time is a constant equal to 3 at the point where the line intercepts the hyperbola. This is for the "horizontal hyperbola". For the vertical hyperbola, \Delta x = \sqrt{(3^2 + \Delta t^2)} the 3 represents the constant proper distance, but imaginary quantities are involved and things get complex (pun intended ). Start with the Minkowski relationship (c\tau)^2 = (ct^2) - x^2 and multiply by both sides by minus one to obtain -(c\tau)^2 = -(ct^2) + x^2 then we get ic\tau = \sqrt{(x^2-(ct)^2)}. For real velocities, where dt>dx the quantity on the left is imaginary, but it is easy to see that for imaginary velocities greater than the speed of light, dt<dx, the proper time is imaginary and the quantity on the left becomes real and is equal to the proper distance, dx_o = ic\tau. Now if we plot lines from the origin, with slopes of dt/dx rather than the usual dx/dt, they intercept the vertical hyperbola at points that all have the same constant proper distance measured by a particle traveling with arbitary velocity v = dx/dt. The reason the imaginary quantities appear is that a proper time interval is the interval between two timelike events measured by a real clock traveling inertially between the two events at less than the speed of light, while the proper distance is measurement of the spacelike interval that is measured by an imaginary clock traveling at greater than the speed of light. Are you confused? You should be. I probably confused myself. Told you things would get complicated Anyway, for the vertical hyperbola, the constant parameter is the real constant proper distance.

 

[stevmg]

You can make any of those variables into a constant. If you draw a line from the origin that intercepts the hyperbola \Delta t = \sqrt{(3^2 + \Delta x^2)} then for ANY velocity \Delta x/ \Delta t in the range (-c<v<c) the elapsed proper time is a constant equal to 3 at the point where the line intercepts the hyperbola. This is for the "horizontal hyperbola". For the vertical hyperbola, \Delta x = \sqrt{(3^2 + \Delta t^2)} the 3 represents the constant proper distance, but imaginary quantities are involved and things get complex (pun intended ). Start with the Minkowski relationship (c\tau)^2 = (ct^2) - x^2 and multiply by both sides by minus one to obtain -(c\tau)^2 = -(ct^2) + x^2 then we get ic\tau = \sqrt{(x^2-(ct)^2)}. For real velocities, where dt>dx the quantity on the left is imaginary, but it is easy to see that for imaginary velocities greater than the speed of light, dt<dx, the proper time is imaginary and the quantity on the left becomes real and is equal to the proper distance, dx_o = ic\tau. Now if we plot lines from the origin, with slopes of dt/dx rather than the usual dx/dt, they intercept the vertical hyperbola at points that all have the same constant proper distance measured by a particle traveling with arbitary velocity v = dx/dt. The reason the imaginary quantities appear is that a proper time interval is the interval between two timelike events measured by a real clock traveling inertially between the two events at less than the speed of light, while the proper distance is measurement of the spacelike interval that is measured by an imaginary clock traveling at greater than the speed of light. Are you confused? You should be. I probably confused myself. Told you things would get complicated Anyway, for the vertical hyperbola, the constant parameter is the real constant proper distance.

You can make any of those variables into a constant. If you draw a line from the origin that intercepts the hyperbola \Delta t = \sqrt{(3^2 + \Delta x^2)} then for ANY velocity \Delta x/ \Delta t in the range (-c<v<c) the elapsed proper time is a constant equal to 3 at the point where the line intercepts the hyperbola.

yiuop, why the "3?"

For the vertical hyperbola, \Delta x = \sqrt{(3^2 + \Delta t^2)} the 3 represents the constant proper distance, but imaginary quantities are involved and things get complex (pun intended ). Start with the Minkowski relationship (c\tau)^2 = (ct^2) - x^2 and multiply by both sides by minus one to obtain -(c\tau)^2 = -(ct^2) + x^2 then we get ic\tau = \sqrt{(x^2-(ct)^2)}. For real velocities, where dt>dx the quantity on the left is imaginary, but it is easy to see that for imaginary velocities greater than the speed of light, dt<dx, the proper time is imaginary and the quantity on the left becomes real and is equal to the proper distance, dx_o = ic\tau. Now if we plot lines from the origin, with slopes of dt/dx rather than the usual dx/dt, they intercept the vertical hyperbola at points that all have the same constant proper distance measured by a particle traveling with arbitary velocity v = dx/dt. The reason the imaginary quantities appear is that a proper time interval is the interval between two timelike events measured by a real clock traveling inertially between the two events at less than the speed of light, while the proper distance is measurement of the spacelike interval that is measured by an imaginary clock traveling at greater than the speed of light. Are you confused? You should be. I probably confused myself. Told you things would get complicated Anyway, for the vertical hyperbola, the constant parameter is the real constant proper distance.

I thought of imaginary or complex numbers and remembered their additive qualities but that pertained to mulitplying. To wit:

Take the value "-1." In complex numbers that's (1)*(sin \pi + i cos \pi) = -1

The \sqrt {-1} = 1*(sin (\pi/2) + i cos (\pi/2) = 0 + i = i

Adding the two angles is for multiplication rather than addition.

 

[stevmg]

Physically the "proper time" along a given worldline just means the time elapsed on a clock that travels along that worldline. If you want to know the proper time between two events on the worldline of a clock moving at constant velocity along the x-axis of some frame, it's just \Delta \tau = \sqrt{\Delta t^2 - (1/c^2) \Delta x^2 }, where \Delta t is the difference in coordinate time between the two events and \Delta x is the difference in coordinate position. This can be rewritten as \Delta \tau = \Delta t \sqrt{1 - (1/c^2) (\Delta x / \Delta t )^2 } = \Delta t \sqrt{1 - v^2 /c^2 } which is just the standard time dilation formula. For a worldline where the velocity is varying in a continuous way, you can approximate its worldline by a "polygonal" worldline made up of a series of constant-velocity segments each lasting a time interval of \Delta t with instantaneous accelerations between them, then if the velocity during the first time interval v₁ (which could just be the average velocity during the same interval for the path with continuously changing velocity that you are trying to approximate), the second has velocity v₂, and the final segment has velocity v_N, then the total elapsed time would just be the sum of the elapsed time on each segment, or \Delta t \sqrt{1 - v_1^2 /c^2} + \Delta t \sqrt{1 - v_2^2 / c^2} + ... + \Delta t \sqrt{1 - v_N^2 /c^2}. And in the limit as as the time interval of each segment goes to zero (so the number of segments approaches infinity), this approximation should approach perfect agreement with the proper time on the original path with continuously-changing velocity. Since an integral is just the limiting case of a sum with smaller and smaller intervals (or an infinite series of intervals which each have an 'infinitesimal' time dt), that means the proper time along a worldline where the velocity as a function of time is given by v(t) can always be computed according to the integral \int_{t_0}^{t_1} \sqrt{1 - v(t)^2 /c^2} \, dt

JesseM -
That is succinct explanation of proper time. Thank you. Now
a) what is a similar explanation for proper velocity?
b) for proper acceleration?

 

[stevmg]

starthaus -

Good. What happens when the proper acceleration a_p increases?

-(ct)^2+x^2=(\frac{c^2}{a_p})^2
should be added to your blog of "Acceleration in SR - II.pdf" as equation 16 as you do have much occasion to refer to it in subsequent discussions.

Also, I relooked at this same .pdf and realized that you were just dealing with an acceleration mode, i.e., where one has an inertial time frame S and a (constant) accelerating time frame S'. Thus your equations pertain to only that and not to two time frames (say, O and O') in which there is constant velocity of O' from O.

S-o-r-r-y... Brain dead again. (I'll rename myself "Fred" so I may be referred to as "Brain Dead Fred.")

stevmg

 

[starthaus]

starthaus -



-(ct)^2+x^2=(\frac{c^2}{a_p})^2
should be added to your blog of "Acceleration in SR - II.pdf" as equation 16 as you do have much occasion to refer to it in subsequent discussions.

Also, I relooked at this same .pdf and realized that you were just dealing with an acceleration mode, i.e., where one has an inertial time frame S and a (constant) accelerating time frame S'. Thus your equations pertain to only that and not to two time frames (say, O and O') in which there is constant velocity of O' from O.

S-o-r-r-y... Brain dead again. (I'll rename myself "Fred" so I may be referred to as "Brain Dead Fred.")

stevmg

Hence the name of the blog file, "Accelerated Motion in SR" :-)

 

[stevmg]

proper time = time elapsed by the inertial observer (must use time-dilation formula or \Delta \tau^2 = \Delta t^2 - \Delta x^2/c^2) = \Delta t^2/\gamma)

What's proper velocity?

What is proper acceleration? (?acceleration as measured by a co-moving inertial ('steady state") observer going at the same instantaneous velocity as the accelerating particle?)

 

[DrGreg]

What's proper velocity?

What is proper acceleration? (?acceleration as measured by a co-moving inertial ('steady state") observer going at the same instantaneous velocity as the accelerating particle?)

See post #11

 

[starthaus]

proper time = time elapsed by the inertial observer (must use time-dilation formula or \Delta \tau^2 = \Delta t^2 - \Delta x^2/c^2) = \Delta t^2/\gamma)

What's proper velocity?

What is proper acceleration? (?acceleration as measured by a co-moving inertial ('steady state") observer going at the same instantaneous velocity as the accelerating particle?)

From the blog:1. Proper speed

v_p=\frac{dx}{d\tau}=\gamma \frac{dx}{dt}=\gamma v2. Proper acceleration

a_p=c \frac{d\phi}{d\tau}=\gamma^3*\frac{dv}{dt}=\gamma^3*\frac{d^2 x}{dt^2}=\gamma^3 a

t=coordinate time
\tau=proper time
v=coordinate speed
a=coordinate acceleration

 

[yuiop]

yiuop, why the "3?"

3 is just a random number that I chose for the constant, but I could have picked any other number. If I had put a letter such as k or n or represent a constant, some people get confused and think I mean a variable. There are various sorts of constants. Some are physical constants are a particular number. Others are sometimes called constants because they are invariant under a transformation and some are constants with respect to time. I meant the last kind. You can choose any number to be the constant (so in that sense it is a variable or a parameter) but once you have chosen it, it remains fixed over time for a given equation. Believe it or not, we have had a long argument on this forum, where someone used the fact that I had used a letter to represent a constant, as "proof" that the constant I using was in fact a variable. LOL

 

[stevmg]

If you are working in a (t,x) coordinate system, and \tau is proper time:

coordinate acceleration = d²x/dt²

proper acceleration = acceleration measured in the coordinate system of a comoving inertial observer = what an accelerometer measures

rapidity = \tanh^{-1} \frac {dx/dt}{c}

coordinate time = t

coordinate velocity = dx/dt

proper velocity = dx/d\tau although I prefer to call it "celerity" because of possible confusions that can occur (especially over proper acceleration).

Thanks Dr Greg, starthaus.

One more thing:

What is proper distance? (I assume this would only apply to events that are spacelike in relationship.)

stevmg

 

[stevmg]

If proper speed (or velocity) is \tau = v*\gamma then it is conceivable that \tau exceed the speed of light. Is that true? If so, what physical meaning does that have? Is there a limit on v_p &lt; c?. Seems like there should be in the sense that there should be NO FR for a given observation in which v \geq c

Likewise, proper acceleration or a_p = a\gamma^3 also can be &gt; c but I have been told that there is no limit on acceleration other than that the coordinate velocity be &lt; c. Is that valid?

starthaus -

How do you create your blog with the .pdf s? To wit, your 3-page MMX.pdf. You obvously create the original in some program or do you use the PF "reply - advanced" box and then delete that box? You can print the LATEX formatted "reply" as a .pdf and then delete (or navigate away from an unposted reply.) Is there another platform which you can use to see LATEX formatted characters?

Don't worry, I am not going to start a blog and if I do, it would be for brain-dead dummies like me (you know, "advanced physics for idiots.pdfs.") There are things I would like to ".pdf-alize" and would like to know how.

stevmg